Question

Differentiate the following w.r.t. x :

(sin x)^tanx + (cos x)^cotx 

Answer 1

Let y = (sin x)^tanx + (cos x)^cotx 
Put u = (sin x)^tanx and v = (cos x)^cotx
Then y = u + v
∴ `"dy"/"dx" = "du"/"dx" + "dv"/"dx"`       ...(1)
Take u = (sin x)^tanx 
∴ log u = log(sin x)^tanx = (tan x).(log sinx)
Differentiating both sides w.r.t. x, we get
`1/u."du"/"dx" = "d"/"dx"[(tan x)(log sin x)]`

= `(tan x)."d"/"dx"(log sin x) + (log sinx)."d"/"dx"(tanx)`

= `(tanx)/(sin x)."d"/"dx"(sin x) + (log sinx)(sec^2x)`

= `((sinx)/(cosx))/(sinx).cosx + (sec^2x)(log sinx)`
= 1 + (sec²x)(log sinx)
∴ `"du"/"dx" = y[1 + (sec^2x)(log sinx)]`

= (sin x)^tanx[1 + (sec²x)(log sinx)]     ...(2)
Also, v = (cos x)^cotx 
∴ log v = log(cos x)^cotx = (cot x).(log cosx)
Differentiating both sides w.r.t. x, we get
`1/v."dv"/"dx" = "d"/"dx"[(cot x).(log cos x)]`

= `(cot x)."d"/"dx"(log cos x) + (log cos x)."d"/"dx"(cotx)`

= `cot x xx 1/cosx."d"/"dx"(cosx) + (log cosx).(-"cosec"^2x)`

= `cotx xx 1/cosx xx (-sin x) - ("cosec"^2x)(log cosx)`

∴ `"dv"/"dx" = v[1/tanx xx (-tanx) - ("cosec"^2x)(log cosx)]`
= –(cos x)^cotx [1 + (cosec²x)(log cosx)]    ...(3)
From (1), (2) and (3), we get
`"dy"/"dx" = (sin x)^(tanx)[1 + (sec^2x)(log sin x)] - (cos x)^(cotx)[1 + ("cosec"^2x)(log cosx)]`.